Problem 92 - Chemistry and Chemical Reactivity

Question

Aqueous solutions of iron(II) chloride and sodium sulfide react to form iron(II) sulfide and sodium chloride. (a) Write the balanced equation for the reaction. (b) If you combine \(40 .\) g each of \(\mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{FeCl}_{2}\) what is the limiting reactant? (c) What mass of FeS is produced? (d) What mass of \(\mathrm{Na}_{2} \mathrm{S}\) or \(\mathrm{FeCl}_{2}\) remains after the reaction? (e) What mass of \(\mathrm{FeCl}_{2}\) is required to react completely with \(40 .\) g of \(\mathrm{Na}_{2} \mathrm{S}\) ?

Step-by-Step Solution

Answer

(a) FeCl₂ + Na₂S → FeS + 2NaCl. (b) FeCl₂ is the limiting reactant. (c) 27.52 g of FeS is produced. (d) 15.61 g of Na₂S remains. (e) 65.55 g of FeCl₂ is required.

Step 1: Write the Balanced Equation

Identify the reactants and products involved in the chemical reaction. The reactants are iron(II) chloride (FeCl₂) and sodium sulfide (Na₂S), and the products are iron(II) sulfide (FeS) and sodium chloride (NaCl). The balanced chemical equation is: \[\text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + 2\text{NaCl}\]

Step 2: Determine Molar Masses

Calculate the molar masses of each compound: - FeCl₂: (55.85 + 2(35.45)) g/mol = 127.75 g/mol - Na₂S: (2(22.99) + 32.07) g/mol = 78.05 g/mol - FeS: (55.85 + 32.07) g/mol = 87.92 g/mol

Step 3: Calculate Moles of Reactants

Find the number of moles of each reactant: - Moles of FeCl₂ = \(\frac{40}{127.75} = 0.313\) mol - Moles of Na₂S = \(\frac{40}{78.05} = 0.513\) mol

Step 4: Identify the Limiting Reactant

Determine the limiting reactant by comparing the stoichiometric amounts from the balanced equation: - From the equation, 1 mol of FeCl₂ reacts with 1 mol of Na₂S. - Since 0.313 mol FeCl₂ < 0.513 mol Na₂S, FeCl₂ is the limiting reactant.

Step 5: Calculate Mass of FeS Produced

Using the moles of limiting reactant (FeCl₂), calculate the moles and then mass of FeS produced:- Moles of FeS = Moles of FeCl₂ = 0.313 mol- Mass of FeS = \(0.313 \times 87.92 = 27.52\) g

Step 6: Calculate Remaining Reactant Mass

Determine the remaining amount of Na₂S. Since Na₂S is in excess:- Moles of Na₂S used = Moles of FeCl₂ = 0.313 mol- Remaining moles of Na₂S = 0.513 - 0.313 = 0.200 mol- Mass of remaining Na₂S = \(0.200 \times 78.05 = 15.61\) g

Step 7: Calculate FeCl₂ for 40g Na₂S

To find the mass of FeCl₂ required to completely react with 40 g of Na₂S, use:- 40 g Na₂S corresponds to 0.513 mol- Moles of FeCl₂ needed = 0.513 mol (from stoichiometry)- Mass of FeCl₂ = \(0.513 \times 127.75 = 65.55\) g

Key Concepts

Balanced Equation

A balanced chemical equation is crucial to understand any chemical reaction. For the reaction between iron(II) chloride (FeCl₂) and sodium sulfide (Na₂S), the balanced equation aligns the quantities of each element on both sides of the reaction. This ensures that the law of conservation of mass is followed, which states that matter cannot be created or destroyed in a chemical reaction.

The goal is to have the same number of each type of atom in the reactants as in the products. In our case, the balanced equation is:\[\text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + 2\text{NaCl}\]From this equation:

• 1 mole of FeCl₂ reacts with 1 mole of Na₂S
• produces 1 mole of FeS and 2 moles of NaCl

The balancing process involves adding coefficients to the chemical formulas to balance the atoms on each side, ensuring the same number appear both as reactants and as products. This is essential for accurately determining stoichiometric relationships in the reaction.

Molar Mass

Molar mass is one of the key concepts for converting between grams and moles in chemical reactions. Every chemical compound has a molar mass, which is the mass of one mole of that compound, measured in grams per mole (g/mol). Understanding molar mass allows you to relate the quantity of a substance in terms of moles, which is essential in stoichiometry.

For our reaction, the molar masses are calculated as follows:

• FeCl₂: (55.85 + 2(35.45)) = 127.75 g/mol
• Na₂S: (2(22.99) + 32.07) = 78.05 g/mol
• FeS: (55.85 + 32.07) = 87.92 g/mol

Knowing these molar masses aids in converting between grams of the substances and moles, which is pivotal when determining the limiting reactant and calculating how much product will form. Use the formula:\[\text{Moles of compound} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\]This formula helps convert the given mass of a reactant to moles, the standard unit of measure for quantity in chemistry.

Chemical Reaction Stoichiometry

Stoichiometry in chemical reactions involves using a balanced equation to calculate the quantities of reactants and products. It allows chemists to predict how much product will be formed from a given amount of reactant, as well as identify the limiting reactant. The limiting reactant is the substance that is entirely consumed first, stopping further reaction.

For the reaction between FeCl₂ and Na₂S, stoichiometry shows that 1 mole of FeCl₂ will react with 1 mole Na₂S, following a 1:1 ratio. From our problem, we have:

• 0.313 mol of FeCl₂
• 0.513 mol of Na₂S

Since you have fewer moles of FeCl₂, it is the limiting reactant. This means that the amount of FeCl₂ will determine the maximum amount of FeS that can be produced. The excess reactant, Na₂S, will have some of its original mass left unreacted.

Using stoichiometry:

• Identify the limiting reactant using the balanced equation's ratios
• Calculate the moles of product formed based on the moles of limiting reactant
• Determine the remaining mass of the excess reactant

Such detailed analysis is key to determining quantities precisely in chemistry.